On perfect order subsets in finite groups Nguyen Trong Tuan High School for gifted VNU - HCM City 145 Nguyen Chi Thanh str., Dist. 5, Ho Chi Minh City, Vietnam
arXiv:1007.0568v1 [math.GR] 4 Jul 2010
e-mail:
[email protected]
Bui Xuan Hai∗ Faculty of Mathematics & Computer Science, University of Science VNU - HCM City 227 Nguyen Van Cu str., Dist. 5, Ho Chi Minh City, Vietnam e-mail:
[email protected]
23 June 2010
Abstract If G is a finite group and x ∈ G then the set of all elements of G having the same order as x is called an order subset of G determined by x (see [2]). We say that G is a group with perfect order subsets or briefly, G is a P OS-group if the number of elements in each order subset of G is a divisor of |G|. In this paper we prove that for any n ≥ 4, the symmetric group Sn is not P OS-group. This gives the positive answer to one of two questions rising from Conjecture 5.2 in [3].
Key words: perfect order subset, finite groups. Mathematics Subject Classification 2010: 20D60
∗
corresponding author
1
1
Introduction
Throughout this paper, all considered groups are finite and for a group G we denote by |G| the order of G, while for an element x ∈ G, the order of x is denoted by o(x). We denote also by N and Z+ the sets of all non-negative and positive integers respectively. If m ∈ Z+ then Gm denotes the direct product G × G × . . . × G. In a group G, define the following equivalence relation: x ∼ y ⇐⇒ o(x) = o(y). The equivalence class defined by an element x is denoted by x and is called an order subset of G. Following the work [2], we say that G is a group with perfect order subsets or briefly, G is a P OS-group if the number of elements in each order subset of G is a divisor of |G|. In [2], the authors study properties of some abelian P OS-groups and they established some curious connection of such groups and Fermat numbers. In [3], the authors have extended their study for non-abelian groups and they have obtained some interesting properties for such groups. Also, in this work, some examples of non-abelian P OS-groups are given. In particular, it is obvious that the symmetric group S3 on three letters is a non-abelian P OS-group. However, the authors conjectured that for n ≥ 4, An and Sn do not have perfect order subsets, i.e. they are not P OS-groups. Recently, in [1], Ashish Kumar Das have proved this conjecture for groups An . Our main purpose in this paper is to prove that the conjecture for groups Sn is also true. As an useful additional information, in Section 2 we give some examples of groups having no perfect order subsets.
2
Examples of groups having no perfect order subsets
In this section we give some examples of groups not necessarily abelian, having no perfect order subsets. In the first, we note that the direct product of P OS-groups does not necessarily be a P OS-group as the following proposition shows. Proposition 2.1 For α, t ∈ Z+ , t > 1, (Z2α )t is not a P OS-group. Proof. The order of any element of (Z2α )t is of the form 2i , i ≤ αt. By [2, Lemma 1], the number of elements of the order 2α is (2α−1 )t (2t − 1). Since t > 1, (2t − 1) does not divide 2αt . Therefore (Z2α )t is not a P OS-group. For 2n ≥ 4, denote by D2n the dihedral group, defined by the following: D2n := ha, b|an = 1, b2 = 1, bab−1 = a−1 i. Lemma 2.1 If n is an even integer then D2n is not a P OS-group. Proof. In a dihedral group D2n := ha, b|an = 1, bn = 1, aba−1 = a−1 i, for every i, 0 ≤ i < n, we have (ai b)2 = 1. So D2n contains n elements of order 2 of this form. Now, if n is even, then an/2 is also an element of order 2. So the number of elements of order 2 is not a divisor of 2n and it follows that D2n is not a P OS-group.
2
Lemma 2.2 If there exist at least two odd prime divisors of n then D2n is not a P OSgroup. Q k , r ≥ 2, pk are all odd primes. Then, Proof. Suppose that n = rk=1 pαkQ Qr the number of r αk −1 elements of Q the order n is ϕ(n) = k=1 pk (pk − 1). It follows that k=1 (pk − 1) is a divisor of 2 rk=1 pk . But, by our assumption this is a contradiction. Theorem 2.1 D2n is a P OS-group if and only if n = 3α , α ∈ Z+ .
Proof. Suppose that D2n is a P OS-group. In view of lemmas 2.1 and 2.2, n = pα for some odd prime p. Then bp = b 6= 1. It follows that every element of a order pα is of the form ai with (i, p) = 1 and 1 ≤ i < pα . So, the number of elements of a order pα is ϕ(pα ) = (p − 1)pα−1 . Since this number is a divisor of 2pα , it follows p = 3. Conversely, suppose that n = 3α . Then, the order of any element of D2n is of the form 2i .3β , where β ≤ α and i ∈ {0, 1}. If i = 0 then the number of elements of a order 3β is ϕ(3β ) = 2.3β−1 , which is a divisor of n = 2.3α . Now, if i = 1 then any element of a order 2.3β must be of the form ak b, 0 ≤ k < n. Then we have (ak b)2 = ak bak b = ak bak b−1 = ak (bak b−1 ) = ak a−k = 1. It follows that β = 0. Thus, such elements have the order 2 and there are exactly n such elements. Hence D2n is a P OS-group. Recall that for n ≥ 3, the generalized quaternion group Qn is defined by the following: n−1
Qn := ha, b|a2
n−2
= 1, b2 = a2
, bab−1 = a−1 i.
Generalized quaternion groups are non-abelian non- P OS groups. In fact, we have the following result: Proposition 2.2 For n ≥ 3, a generalized quaternion group Qn is not a P OS-group. Proof. Consider a generalized quaternion group n−1
Qn := ha, b|a2
n−2
= 1, b2 = a2
, bab−1 = a−1 i.
Since bab−1 = a−1 , bai b−1 = a−i for all i, 0 ≤ i < 2n−1 . From the last equality it follows n−2
(ai b)2 = ai (bai b−1 )b2 = ai a−i b2 = b2 = a2 n−2
n−1
.
Hence (ai b)4 = (a2 )2 = a2 = 1. So, there are 2n−1 elements of the order 4 of the n−3 form ai b, 0 ≤ i < 2n−1 . On the other hand, the order of the element a2 is 4. Hence, the number of elements of order 4 does not divide 2n . So Qn is not P OS-group.
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3
The symmetric groups
As we have mentioned in the Introduction, the symmetric group S3 is a P OS-group. In [3, Conjecture 5.2], the authors conjectured that for any n ≥ 4, the group Sn is not a P OS-group. Our main purpose in this section is to give the positive answer to this conjecture. In fact, we shall prove the following result: Theorem 3.1 For any integer n ≥ 4, the symmetric group Sn is not a P OS-group. To prove this theorem, we need some lemmas. Lemma 3.1 Let p be an odd prime number. If n = 2p + r with r ∈ {0, 1, . . . , p − 1}, then Sn is not a P OS-group. Proof. Suppose that under our supposition, Sn is a P OS-group. Consider any element α of the order p in Sn . Then, either α is a cycle of the length p or it is a product of two disjoint cycles of the same length p. For the convenience, we call α an element of type 1 for the first case and α an element of type 2 for the second case. Obviously that the number of all elements of type 1 is n! n! Apn = = p p(n − p)! p(p + r)! and the number of all elements of type 2 is p 1 Apn An−p 1 n! (n − p)! n! × × = × × = 2 . 2 p p 2 p(n − p)! p(n − 2p)! 2p r!
Hence, the number of all elements of the order p in Sn is d=
n! 2pr! + (p + r)! n! + 2 = n! p(p + r)! 2p r! 2p2 r!(p + r)!
Since Sn is a P OS-group,
2p2 r!(p + r)! 2pr! + (p + r)!
is an integer or 2p3 r!(r + 1) . . . (p − 1)(p + 1) . . . (p + r) 2p2 (p + r)! = 2p + (r + 1)(r + 2) . . . (p + r) p[2 + (r + 1) . . . (p − 1)(p + 1) . . . (p + r)] is an integer. Therefore 2p2 r!(r + 1) . . . (p − 1)(p + 1) . . . (p + r) 2 + (r + 1) . . . (p − 1)(p + 1) . . . (p + r) is an integer. Since (r + 1) . . . (p − 1)(p + 1) . . . (p + r) ≡ (p − 1)! (mod p), 4
(1)
in view of Wilson’s Theorem we have 2 + (r + 1) . . . (p − 1)(p + 1) . . . (p + r) ≡ 1
(mod p).
(2)
From (1) and (2) it follows that 2r!A 2r!(r + 1) . . . (p − 1)(p + 1) . . . (p + r) = 2 + (r + 1) . . . (p − 1)(p + 1) . . . (p + r) 2+A is an integer, where A = (r + 1) . . . (p − 1)(p + 1) . . . (p + r). Since gcd(A, 2 + A) = 4r! gcd(A, 2) = 1 or 2, is an integer, that is a contradiction in view of the following 2+A inequalities: 2 + A > (p + 1)(p + 2)(p + 3) . . . (p + r) > 4(p + 2) . . . (p + r) > 4(1.2 . . . r) = 4r!. The proof is now complete. Lemma 3.2 If n = 3p + r, where p is a odd prime and r ∈ {0, 1, 2, . . . , p − 1}, then Sn is not a P OS-group. Proof. Suppose that under our supposition, Sn is a P OS-group. Consider any element α of the order p in Sn . Then, either α is a cycle of the length p or it is a product of two or three disjoint cycles of the same length p. The number of elements in each of these cases is n! n! Apn = = , p p(n − p)! p(2p + r)! p 1 Apn An−p 1 n! (n − p)! n! × × = × × = 2 2 p p 2 p(n − p)! p(n − 2p)! 2p (p + r)!
and p p 1 n! (n − p)! (n − 2p)! n! 1 Apn An−p An−2p × × × = × × × = 3 6 p p p 6 p(n − p)! p(n − 2p)! p(n − 3p)! 6p r!
respectively. Hence, the number of elements of the order p in Sn is 1 1 1 d = n! . + + p(2p + r)! 2p2 (p + r)! 6p3 r! Since Sn is a P OS-group, d must be divided n! and, consequently k=
6p3 .r!(p + r)!(2p + r)! 6p2 (p + r)!r! + 3p(2p + r)!r! + (2p + r)!(p + r)!
is an integer. By setting A := (r + p + 1) . . . (2p − 1)(2p + 1) . . . (2p + r) and the direct calculation we have k=
6p3 .(p − 1)!(p + 1) . . . (p + r)A . 3 + 3A + (r + 1) . . . (p − 1)(p + 1) . . . (p + r)A 5
(3)
By applying of Wilson’s Theorem we get (r + 1) . . . (p − 1)(p + 1) . . . (p + r) ≡ −1
(mod p)
(4)
and, consequently we have A = (p + r + 1) . . . (2p − 1)(2p + 1) . . . (2p + r) ≡ −1
(mod p).
(5)
From (4) and (5) it follows that 3 + 3A + (r + 1) . . . (p − 1)(p + 1) . . . (p + r)A ≡ 1 (mod p). Since gcd (A, 3 + A[3 + (r + 1) . . . (p − 1)(p + 1) . . . (p + r)]) = gcd(3, A) and k is an integer, it follows from (3) that B :=
18(p − 1)!(p + 1) . . . (p + r) 3 + 3A + (r + 1) . . . (p − 1)(p + 1) . . . (p + r)A
is an integer. Now, we claim that (r + 1) . . . (p − 1)(p + 1) . . . (p + r)A > 18(p − 1)!(p + 1) . . . (p + r). In fact, this inequality is equivalent to the following one: A = (r + p + 1) . . . (2p − 1)(2p + 1) . . . (2p + r) > 18r!. Since p is an odd prime, the last inequality holds as the following calculation shows: A = (2p + 1)(2p + 2) . . . (2p + r)(r + p + 1) . . . (2p − 1) ≥ (2p + 1)(p + 1)(2.3 . . . r) = (2p + 1)(p + 1).r! > 18r!. Clearly, what we have claimed shows that B is not an integer. This contradiction completes the proof of the lemma. Lemma 3.3 If n = 4p, where p is a odd prime, then Sn is not a P OS-group. Proof. Suppose that under our supposition, Sn is a P OS-group. Let d be the number of elements of the order p in Sn . Then we have p p p p p p Ap 1 Ap An−p 1 Apn An−p An−2p 1 Apn An−p An−2p An−3p d= n+ × n× + × × × + × × × × p 2 p p 6 p p p 24 p p p p 1 1 1 1 = n! + 2 + 3 + p(3p)! 2p (2p)! 6p p! 24p4 3 24p .(2p)!p! + 12p2 .(3p)!p! + 4p.(3p)!(2p)! + p!(2p)!(3p)! .n! = 24p4.p!(2p)!(3p)! 24p3 + 12p2 (2p + 1) . . . (3p) + 4p(p + 1) . . . (3p) + (3p)! .n!. = 24p4 (3p)!
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Since d divides n!, 24p4 (3p)! n! = d 24p3 + 12p2 (2p + 1) . . . (3p) + 4p(p + 1) . . . (3p) + (3p)! is an integer. By dividing both numerator and denominator of the last fraction by 6p3 we get n! 24p4 .(p − 1)!(p + 1) . . . (2p − 1)(2p + 1) . . . (3p − 1) = . d 4 + (2p + 1) . . . (3p − 1)[6 + 4(p + 1) . . . (2p − 1) + (p − 1)!(p + 1) . . . (2p − 1)] By setting A = (2p + 1) . . . (3p − 1) and M = 4 + A[6 + 4(p + 1) . . . (2p − 1) + (p − 1)!(p + 1) . . . (2p − 1)], we have
n! 24p4 (p − 1)!(p + 1) . . . (2p − 1)A = . d M In view of Wilson’s Theorem we have
(2p + 1) . . . (3p − 1) = (2p + 1)(2p + 2) . . . (2p + p − 1) ≡ (p − 1)! ≡ −1 (p + 1) . . . (2p − 1) = (p + 1)(p + 2) . . . (p + p − 1) ≡ (p − 1)! ≡ −1
(6)
(mod p); (mod p).
Hence M ≡ 4 + (−1)[6 − 4 + (−1)(−1)] ≡ 1 (mod p). Consequently, gcd(M, p4) = 1 . Therefore, in view of (6) we conclude that 24(p − 1)!(p + 1) . . . (2p − 1)A M is an integer. Note that, if m is a common divisor of A and M, then m must divide 4. In particular, gcd(A, M) must be 1, 2 or 4. It follows that C :=
96(p − 1)!(p + 1) . . . (2p − 1) M
is an integer. However, we can check that C is not an integer for any odd prime p. In fact, if p = 3, then 3840 C= 7060 which is not an integer. Now, suppose that p ≥ 5. Then we have A = (2p + 1) . . . (3p − 1) ≥ (2.5 + 1).12.13.14 > 96 and M > (p − 1)!(p + 1) . . . (2p − 1)A > 96(p − 1)!(p + 1) . . . (2p − 1). Hence, in this case C is not an integer too. This contradiction completes the proof of the lemma. 7
Now, we are ready to prove the main theorem in this section. Proof of Theorem 3.1. For n = 6 and n = 7, the desired result follows from Lemma 3.1 by taking p = 3, r = 0 and p = 3, r = 1 respectively. For n = 4 and n = 5, note that the number of elements of the order 2 in S4 and S5 is 9 and 25 respectively. So, S4 and S5 are both non-P OS groups. hni . According to Bertrand’s Postulate (see, for Now, suppose that n ≥ 8 and m = 4 example [4, Theorem 5.8, p. 109]), there exists some prime p such that m < p < 2m. Note that p < 2m = 2
hni
≤2
n n = . 4 2
4 n < p ≤ , then n = 4p and the conclusion follows from Lemma 3.3. Therefore, If 4 4 we can suppose that n n
